Given an undirected graph, how to check if there is a cycle in the graph? For example, the following graph has a cycle 1-0-2-1.
We have discussed cycle detection for the directed graph. We have also discussed a union-find algorithm for cycle detection in undirected graphs.. The time complexity of the union-find algorithm is O(ELogV). Like directed graphs, we can use DFS to detect a cycle in an undirected graph in O(V+E) time. We have discussed DFS based solution for cycle detection in an undirected graph.
In this article, the BFS based solution is discussed. We do a BFS traversal of the given graph. For every visited vertex ‘v’, if there is an adjacent ‘u’ such that u is already visited and u is not a parent of v, then there is a cycle in the graph. If we don’t find such an adjacent for any vertex, we say that there is no cycle.
We use a parent array to keep track of the parent vertex for a vertex so that we do not consider the visited parent as a cycle.
Time Complexity: The program does a simple BFS Traversal of graph and graph is represented using adjacency list. So the time complexity is O(V+E)
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